Probability Review for Financial Engineers Part 2

Conditional Probability The conditional probability that E occurs given that F has occurred is denoted by 𝑃(𝐸|𝐹) If P(F) > 0 then 𝑃 𝐸 𝐹 = 𝑃(𝐸𝐹) 𝑃(𝐹)

Example – 2 dice 2 dice are rolled - a red dice and a green dice, What is the probability distribution for the total? 2 - 1/36 3 - 2 * (1/36) … 7 – 6 * (1/36) 11 – 2 (1/36) 12 – 1/36

2 Dice example continued   What is the expected value? 7 Ex) What is the probability distribution function for the total given that the Green dice was a 3, that is P(T|G=3) 4 – 1/6 5 – 1/6 6 – 1/6 7 – 1/6 8 – 1/6 9 – 1/6

Example) Playing Cards Selecting a card a standard 52 playing card deck What is the probability of getting an ace? 4/52 = 1/13 What is the probability of getting a ace given that someone already removed a jack from the deck? 4/51 the removal of a jack means that a non-ace has been removed from the deck What is the probability of getting an ace given that someone already removed a spade from the deck? 1/13 the removed cards suit is independent of the rank question.

Joint Cumulative Distributions F(a,b) = P(X≤a, Y≤b) The distribution of X can be obtained from the joint distribution of X and Y as follows 𝐹 𝑋 =𝑃 𝑋<𝑎 =𝑃 𝑋<𝑎|𝑌< ∞ =𝑃( lim 𝑏→∞ 𝑋<𝑎|𝑌< 𝑏 ) = lim 𝑏→∞ 𝑃 𝑋<𝑎|𝑌< 𝑏 = lim 𝑏→∞ 𝐹(𝑎,𝑏) =𝐹(𝑎,∞)

Example – Time between arrivals A market buy order and a market sell order arrive uniformly distributed between 1 and 2pm. Each person puts a 10 minute time limit on each order. What is the probability that the trade will not be executed because of a timeout?   This would be the P(B +10 < S) + P(S+10 < B) = 2 P(B +10 < S) =2 𝐵+10<𝑆 𝑓 𝑏,𝑠 𝑑𝑏 𝑑𝑠 =2 𝐵+10<𝑆 𝑓 𝐵 (𝑏) 𝑓 𝑆 (𝑠)𝑑𝑏 𝑑𝑠 =2 10 60 0 𝑠−10 1 60 2 𝑑𝑏 𝑑𝑠 = 2 60 2 10 60 𝑠−10 𝑑𝑠 = 25 36

Expected Values of Joint Densities Suppose f(x,y) is a joint distribution 𝐸[𝑔 𝑋 ℎ 𝑌 ] = −∞ ∞ −∞ ∞ 𝑔 𝑥 ℎ 𝑥 𝑓(𝑥,𝑦)𝑑𝑥 𝑑𝑦 = −∞ ∞ −∞ ∞ 𝑔 𝑥 ℎ 𝑥 𝑓 𝑋 (𝑥) 𝑓 𝑌 (𝑦)𝑑𝑥 𝑑𝑦 = −∞ ∞ ℎ 𝑥 𝑓 𝑌 (𝑦)𝑑𝑦 −∞ ∞ 𝑔 𝑥 𝑓 𝑋 (𝑥)𝑑𝑥 =𝐸[ℎ 𝑌 ]𝐸[𝑔 𝑋 ]

Covariance of 2 Random Variables 𝐶𝑜𝑣 𝑋,𝑌 =𝐸 (𝑋−𝐸 𝑋 ∗ 𝑌−𝐸 𝑌 ] =𝐸 𝑋𝑌 −𝐸 𝑋 𝑌−𝑋𝐸 𝑌 +𝐸 𝑋 𝐸[𝑌] =𝐸 𝑋𝑌 −𝐸 𝑋 𝐸[𝑌]−𝐸[𝑋]𝐸 𝑌 +𝐸 𝑋 𝐸[𝑌] =𝐸 𝑋𝑌 − 𝐸 𝑋 𝐸[𝑌] Note that is X and Y are independent, then the covariance = 0

Variance of sum of random variables 𝑉𝑎𝑟 𝑋+𝑌 = 𝐸[ 𝑋+𝑌−𝐸 𝑋+𝑌 2 ] = 𝐸[ 𝑋+𝑌−𝐸𝑋−𝐸𝑌 2 ] = 𝐸[ 𝑋−𝐸𝑋+𝑌−𝐸𝑌 2 ] = 𝐸 𝑋−𝐸𝑋 2 + 𝑌−𝐸𝑌 2 +2 𝑋−𝐸𝑋 𝑌−𝐸𝑌 = 𝐸 𝑋−𝐸𝑋 2 ]+ 𝐸[ 𝑌−𝐸𝑌 2 +2𝐸[ 𝑋−𝐸𝑋 𝑌−𝐸𝑌 ] 𝑉𝑎𝑟 𝑋+𝑌 = 𝑉𝑎𝑟 𝑋 +𝑉𝑎𝑟 𝑌 +2𝐶𝑜𝑣(𝑋,𝑌)

Correlation of 2 random variables As long as Var(X) and Var(Y) are both positive, the correlation of X and Y is denotes as 𝜌 𝑋,𝑌 = 𝐶𝑜𝑣(𝑋,𝑌) 𝑉𝑎𝑟 𝑋 𝑉𝑎𝑟(𝑌) It can be shown that −1 ≤ 𝜌 𝑋,𝑌 ≤1 The correlation coefficient is a measure of the degree of linearity between X and Y 𝜌 𝑋,𝑌 =0 means very little linearity 𝜌 𝑋,𝑌 𝑛𝑒𝑎𝑟+1 means X and Y increase and decrease together 𝜌 𝑋,𝑌 𝑛𝑒𝑎𝑟−1 means X and Y increase and decrease inversely

Central Limit Theorem Loosely put, the sum of a large number of independent random variables has a normal distribution. Let 𝑋 1 , 𝑋 2 … be a sequence of independent and identically distributed random variables each having mean 𝜇 and variance 𝜎 2 Then the distribution of 𝑋 1 +…+ 𝑋 𝑛 −n𝜇 𝜎 𝑛 Tends to a standard normal as n ∞, that is 𝑃 𝑋 1 +…+ 𝑋 𝑛 −n𝜇 𝜎 𝑛 ≤𝑎 → 1 2𝜋 −∞ 𝑎 𝑒 − 𝑥 2 /2 𝑑𝑥